What is the difference between dynamics and kinematics

A rod (length = 80 cm) with a rectangular cross section (1.5 mm × 2.0 mm) has a resistance of 0.20 Ω. What is the resistivity of

Andru [333]

Answer:

The resistivity of the material used to make the rod is ρ= 7.5 * 10⁻⁷ Ω.m

Explanation:

R= 0.2 Ω

L= 0.8 m

S= 1.5mm*2mm= 3 mm² = 3 * 10⁻⁶ m²

ρ = (R*S)/L

ρ= 7.5 * 10⁻⁷ Ω.m

7 0

3 years ago

Which is NOT considered as a property of electromagnetic waves: 5

OlgaM077 [116]

Answer:

require a medium to propagate

Explanation:

this is because electromagnetic waves can travel through vacuum at the speed of 3.0 x 10^8 m/s

because they don't need particles to transfer energy

hope this helps

4 0

3 years ago

Why electric field inside the conductor zerqual to zero

Arlecino [84]

In electrostatics free charges in a good conductor<span> reside only on the surface.

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5 0

3 years ago

Figure 8-56 shows a solid, uniform cylinder of mass 7.00 kg and radius 0.450 m with a light string wrapped around it. A 3.00-N t

AVprozaik [17]

Answer:

a) The cylinder has an angular acceleration of 3.810 radians per square second, b) The frictional force has a magnitude of 9 newtons and has the same direction of tension force.

Explanation:

The external force exerted on string creates a tension force that tries to move the cylinder in translation, but it is opposed by the friction force between cylinder and ground that generates rolling on cylinder. The Free Body Motion on cylinder-string system is presented below as attachment. Given that cylinder is a rigid body in planar motion, two equations of equilibrium for translation and an equation of equilibrium for rotation are needed to represent the system, which are now described:

$\Sigma F_{x} = T + f = M\cdot R\cdot \alpha$

$\Sigma F_{y} = N - M\cdot g = 0$

$\Sigma M_{G} = (T-f)\cdot R = I_{G}\cdot \alpha$

Where:

$T$ - Tension, measured in newtons.

$f$ - Friction force, measured in newtons.

$M$ - Mass of the cylinder, measured in kilograms.

$R$ - Radius of the cylinder, measured in meters.

$\alpha$ - Angular acceleration, measured in radians per square second.

$N$ - Normal force from ground exerted on cylinder, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$I_{G}$ - Moment of inertia of the cylinder with respect to its center of mass, measured in kilogram-square meters.

The moment of inertia of the cylinder is:

$I_{G} = \frac{1}{2}\cdot M\cdot R^{2}$

a) The angular acceleration is determined by solving on first and third equation after eliminating friction force:

$f = M\cdot R \cdot \alpha - T$

$(T-M\cdot R\cdot \alpha+T) \cdot R = I_{G}\cdot \alpha$

$2\cdot T\cdot R = (I_{G} + M\cdot R^{2})\cdot \alpha$

$\alpha = \frac{2\cdot T\cdot R}{I_{G}+M\cdot R^{2}}$

$\alpha = \frac{2\cdot T \cdot R}{\frac{1}{2}\cdot M\cdot R^{2}+M\cdot R^{2} }$

$\alpha = \frac{4\cdot T}{3\cdot M\cdot R}$

If $T = 3\,N$ , $M = 7\,kg$ and $R = 0.45\,m$ , then:

$\alpha = \frac{4\cdot (3\,N)}{(7\,kg)\cdot (0.45\,m)}$

$\alpha = 3.810\,\frac{rad}{s^{2}}$

The cylinder has an angular acceleration of 3.810 radians per square second.

b) The magnitude of the frictional force can be determined with the help of the following expression:

$f = M\cdot R \cdot \alpha - T$

Given that $T = 3\,N$ , $M = 7\,kg$ , $R = 0.45\,m$ and $\alpha = 3.810\,\frac{rad}{s^{2}}$ , the magnitude of the friction force is:

$f = (7\,kg)\cdot (0.45\,m)\cdot \left(3.810\,\frac{rad}{s^{2}} \right)-3\,N$

$f = 9\,N$

The frictional force has a magnitude of 9 newtons and has the same direction of tension force.

5 0

4 years ago

A boy pushes on a wagon so that it accelerates at a rate of 0.50 m/s2. The wagon has a mass of 24 kg. What is the magnitude of t

Roman55 [17]

To find the magnitude of the boy's applied force, use newton's second law (where the force is the mass times the acceleration)

F = ?
m = 24 kg
a = 0.50 m/s^2

F = ma
F = (24 kg) (0.50 m/s^2)
F = 12 N

therefore the magnitude of the boy's pushing force is 12 N

6 0

4 years ago

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